116. Populating Next Right Pointers in Each Node

Given a binary tree

struct TreeLinkNode {
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to

NULL

.

Initially, all next pointers are set to

NULL

.

Note:

You may only use constant extra space.

You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,

1
/  \
2    3
/ \  / \
4  5  6  7

After calling your function, the tree should look like:

1 -> NULL
/  \
2 -> 3 -> NULL
/ \  / \
4->5->6->7 -> NULL

时间复杂度：o(ｎ) 空间复杂度：Ｏ（１），因为记录了next指针，可以在层里查找同层的下一个元素，在做层序遍历时不用用到queue,减少了时间复杂度。
机智，即使是相同的办法，在不同的条件下，可以有优化。比如这题里，想想为啥层序遍历要用ｑｕｅｕｅ，为了找到上一层的下一个元素。如果有了ｎｅｘｔ指针，不用ｑｕｅｕｅ也能找到上一层的下一个元素，得到一个优化解法。 ３０％

/**
* Definition for binary tree with next pointer.
* public class TreeLinkNode {
*     int val;
*     TreeLinkNode left, right, next;
*     TreeLinkNode(int x) { val = x; }
* }
*/
public class Solution {
public void connect(TreeLinkNode root) {
if (root == null) {
return;
}
root.next = null;
TreeLinkNode prevLevelStart = root, prev, cur;
while (prevLevelStart.left != null) {
prev = prevLevelStart;
cur = prevLevelStart.left;
while (prev != null) {
if (cur == prev.left) {
cur.next = prev.right;
} else {
prev = prev.next;
if (prev == null) {
cur.next = null;
} else {
cur.next = prev.left;
}
}
cur = cur.next;
}
prevLevelStart = prevLevelStart.left;
}
}
}

相当与层序遍历，在遍历一层时，把一层里的元素前后连起来。 9% 时间＋空间复杂度: O(n)

/**
* Definition for binary tree with next pointer.
* public class TreeLinkNode {
*     int val;
*     TreeLinkNode left, right, next;
*     TreeLinkNode(int x) { val = x; }
* }
*/
public class Solution {
public void connect(TreeLinkNode root) {
ArrayDeque<TreeLinkNode> queue = new ArrayDeque<TreeLinkNode>();
if (root == null) {
return;
}
int size = 0, i = 0;
TreeLinkNode tmp = null;
queue.offer(root);
while (!queue.isEmpty()) {
size = queue.size();
i = 0;
while (i < size) {
tmp = queue.poll();
if (tmp.left != null) {
queue.offer(tmp.left);
}
if (tmp.right != null) {
queue.offer(tmp.right);
}
if (i != size - 1) {
tmp.next = queue.peek();
} else {
tmp.next = null;
}
i++;
}
}
}
}